**What is a Transformer in Physics?**

A transformer is an electrical instrument or device, and it helps us to regulate the size of an alternating current voltage by stepping up or stepping down the voltage. A transformer can increase or decrease the input or output of an alternating current. In this article, you will learn how to calculate the efficiency of a transformer.

## Formula for Efficiency of a Transformer

We can define the efficiency of a transformer as the ratio of total power output to the total power input multiplied by 100%. The efficiency of a transformer symbol is written as η ( known as eta, and it originates from Greek modern letters).

The formula for calculating efficiency of a transformer is

η = (output power / input power) x 100%

Thus, the formula for calculating the efficiency of a transformer is η = (I_{s}E_{s}/I_{p}E_{p}) x 100%

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### How to Calculate the Efficiency of a Transformer

Here is how to calculate the efficiency of a transformer:

#### Problem 1

If the current in the 240-volt supply is 1.25A, and the current in the secondary is 10A. The voltage ratio is also 24:240 volts. Calculate the efficiency of the transformer.

**Solution:**

Current in the primary coil, I_{p} = 1.25A

Also, Current in the secondary coil, I_{s} = 10A

Voltage in the primary coil, E_{p} = 24V

Additionally, the voltage in the secondary coil, E_{s} = 240V

Hence, we apply the formula for the efficiency of the transformer (η = I_{s}E_{s}/I_{p}E_{p})

After substituting our data into the formula for efficiency, we will have

η = (10 x 240) / (1.25 x 24) = 2,400/30 = 80%

**Therefore, the efficiency of the transformer η is 80%**

#### Problem 2

A transformer supplies 15V from 220V mains. If the transformer takes 0.7A From the mains when used to light three lamps connected in parallel and each rated 15V and 40W. Calculate the efficiency of the transformer.

**Solution**

**Data:**

Voltage in the primary coil, V_{p} = 220V

We also have Voltage in the secondary coil, V_{s} = 15V

The current in the primary coil, I_{p} = 0.7A

Current in the secondary coil, I_{s} = ?

We can now apply the formula that says I_{s}V_{s} = I_{p}V_{p} and make I_{s} subject of the formula

The **power output in the primary coil** = I_{p}V_{p} = (0.7 x 220) = 154W

The **power output in the secondary coil** = I_{s}V_{s} = (I_{s} x 15)W

Additionally, we know that current, I = Power (P)/ Voltage(V) = P/V

Thus, I_{s} = 40W / 15V = 2.7A = 3A

Since we have 3 lamps in parallel. Our total current I_{s} will be 3 lamps multiplied by 3-ampere to get 9-ampere.

which implies I_{s} = 3lamps x 3A = 9A

Therefore, our output power = I_{s} x 15 = 9 x 15 = 135W

We can now comfortably apply the formula for efficiency of a transformer that says

Efficiency, η = (output power / input power) x 100% = (135/154) x 100% = 0.88 x 100% = 88%

Therefore, the efficiency of the transformer is 88%

## How Transformer Operates

A transformer consists of a coil, laminated soft iron core, primary coil, secondary coil, the alternating voltage at the primary coil (known as Ep), the alternating voltage at the secondary coil (known as Es), the number of turns in the primary coil (Np), and the number of turns in the secondary coil (N_{s}).

The device is created with a soft iron core, which is wrapped with a coil of wire from both ends. By the left side of a transformer, we have a primary coil that circles the soft iron core.

Similarly, we laminate the soft iron core to avoid loss of energy due to the heat generated by the process. The lamination is also important because it will remove the effect of eddy current. Laminating the iron core can boost the efficiency of the transformer.

Additionally, through the primary coil, an alternating voltage E_{p}, is channeled to the iron core. The iron core will in turn transfer magnetic flux to the other side of the core (right-hand side). The right-hand side of the core is wrapped with another coil, and we call it the secondary coil.

A secondary coil receives a.c voltage of the same frequency as that of a primary coil. Therefore, each coil receives an equal amount of a.c voltage on each side. Thus, we will end up having a magnetic flux that is proportional to the number of turns in both primary and secondary coils.

## Transformer Formula

We will now derive the formulae that can help us understand how to calculate the efficiency of a transformer.

After reading the working principle of a transformer, we can confidently come up with a transformer formula as follows:

The formula for the proportionality of the number of turns to the alternating voltage is written below

E_{s}/E_{p} = N_{s}/N_{p}

[where E_{s} = a.c voltage at the secondary coil, Ep = a.c voltage at the primary coil, N_{s} = Number of turns at the secondary coil, and N_{p} = Number of turns at the primary coil]

Therefore, we can finally say that:

(**a.c voltage** **at the secondary coil** / **a.c voltage at the primary coil**) **=** (**Number of turns at the secondary coil / Number of turns at the primary coil**)

When a transformer is 100% efficient, the power in both coils is the same. Since we know that power is equal to the product of current and voltage, it is safe to say that:

P_{secondary coil} = P_{primary coil}

and

P_{secondary coil} = I_{s}E_{s}

[ where I_{s} = current in the secondary coil, and E_{s} = Voltage in the secondary coil]

P_{primary coil} = I_{p}E_{p}

[ where I_{p} = current in the primary coil, and E_{p} = Voltage in the primary coil]

Thus, I_{s}E_{s} = I_{p}E_{p}

which in turn became E_{s}/E_{p} = I_{p}/I_{s}

Therefore, we can now say

E_{s}/E_{p} = I_{p}/I_{s}

The above equations will help you to understand how to find the efficiency of a transformer

## Step-up and Step-down Transformer

The increase in the output of the alternating voltage from primary to secondary winding, **signifies a step-up transformer**. While **step-dwon transformer** works by drecreasing the output alternating voltage.

The **difference between step-up and step-down transformer** are as follows:

We apply step-up transformer from the generating sites, because we need to send high voltages to far places.For step-down transformer, we need to reduce the high voltages sent from the generating sites, this will enable us to make use of the electricity at home and avoid damages.

## What Causes Energy Loss in a Transformer?

The loss of energy in a transformer can be due to the following factors:

There will be a loss in energy due to the circulation of electric current in an electric conductor which affects the magnetic field in the conductor. The iron core will lose heat energy due to the eddy current.

**2. Hysteresis Loss:**

We lose heat energy in the iron core due to the to and fro pattern of magnetization and demagnitization of the iron core.

3. **Copper Loss:** Lost of heat energy due to I^{2}R

4. When **magnetic flux is leaking**

## How to Reduce the Loss of Energy in a Transformer

Here are ways that can help us to reduce the loss of energy in a transformer:

## Solved Problems on Transformer

We will solve another problem on how to calculate the efficiency of transformer and have a look at few examples and their solutions on transformer.

### Example 1

A step-up transformer operates on a 220V and supplied a current of 10A. The ratio of the primary to secondary winding is 1:10. Calculate the

a. Voltage across the secondary

b. The current in the primary

c. Power output

**Solution**

**Data**

Primary voltage, E_{p} = 220V

Secondary voltage, E_{s} = ?

Primary Current, I_{p} = ?

Secondary Current, I_{s} = 10A

N_{p}/N_{s} = 1/10

#### Solution a

We apply the formula that says E_{p}/E_{s} = N_{p}/N_{s}

By substituting our data into the above formula, we will get

220/E_{s} = 1/10

and when we make E_{s} subject of the formula, we will get

E_{s} = 220 x 10 = 2,200V

Therefore, the voltage across the secondary is 2,200V

#### Solution b

We apply the formula that says E_{s}I_{s} = E_{p}I_{p}

Thus, by making I_{p} subject of the formula, we will get

I_{p} = E_{s}I_{s}/E_{p} = (2,200 x 10)/220 = 22,000/220 = 100A

Therefore, the current at the primary is 10-Ampere

#### Solution c

Power output = E_{s}I_{s} = 2,200 x 10 = 22,000W = 22kW

Therefore, the power output is 22-kilowatt

### Example 2

A house is supplied with a 240V a.c mains. To operate a door bell rated at 8V, a transformer is used. If the number of the turns in the primary coil of the transformer is 900. Calculate the number of turns in the secondary coil of the transformer.

**Solution**

**Data:**

Primary voltage, E_{p} = 240V

Secondary voltage, E_{s}= 8V

Primary number of turns, N_{p} = 900

Secondary number of turns, N_{s} = ?

By applying the formula E_{s}/E_{p} = N_{s}/N_{p}

and making N_{s} subject of the formula

N_{s} = E_{s}N_{p} / E_{p} = (8 x 900) / 240 = 30

### Example 3

If a transformer is used to light a lamp at 60W, 220V from a 4,400V a.c supply, calculate the

a. Turn ratio of the trnasformer

b. Current taken from the main ciruit, if the efficiency of the trnasformer is 95%

**Solution**

**Data**

Primary voltage, E_{p} = 4,400V

Secondary voltage, E_{s} = 220V

Output power, P_{s} = I_{s}E_{s} = 60W

Efficiency of the transformer, η = 95% = 95/100 = 0.95

#### Solution a

We can use the formula that says E_{p}/E_{s} = N_{p}/N_{s} to find the turn ratio of the transformer

where N_{s} / N_{p} = 220/4,400 = 1/20

Therefore the turn ratio of the transformer is N_{s}**:**N_{p} = 1**:**20

#### Solution b

We can apply the formula for the efficiency of the transformer (η = I_{s}E_{s}/I_{p}E_{p}) x 100% to solve this problem

Therefore, after inserting our data into the above formula, we will have

0.95 = 60/ (I_{p} x 4,400)

We now make I_{p} subject of the formula

I_{p} = 60 / (0.95 x 4,400) = 60/4,180 = 0.014354A

Therefore, the current is 0.014-Ampere

With the above detail explanation on transformer and how to calculate the effieciency of transformer. I hope you will be able to solve problems on transformer by yourself?

Drop a comment if there is any part of the topic that you dont understand.

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